Combine the integrals:
\frac{1}{2} \ln\left|\frac{1 + y}{1 - y}\right| = -\frac{1}{t} + C
Multiply through by 2:
\ln\left|\frac{1 + y}{1 - y}\right| = -\frac{2}{t} + 2C
Exponentiate both sides to remove the logarithm:
\left|\frac{1 + y}{1 - y}\right| = e^{-\frac{2}{t} + 2C}
Let A = e^{2C}, which simplifies the equation to:
\frac{1 + y}{1 - y} = A e^{-\frac{2}{t}}
Solve for y:
Cross-multiply:
1 + y = (1 - y) A e^{-\frac{2}{t}}
Expand the right-hand side:
1 + y = A e^{-\frac{2}{t}} - A e^{-\frac{2}{t}} y
Now gather terms involving y on one side:
y + A e^{-\frac{2}{t}} y = A e^{-\frac{2}{t}} - 1
Factor out y on the left-hand side:
y(1 + A e^{-\frac{2}{t}}) = A e^{-\frac{2}{t}} - 1
Solve for y:
y = \frac{A e^{-\frac{2}{t}} - 1}{1 + A e^{-\frac{2}{t}}}
Final Answer:
The general solution is:
y = \frac{A e^{-\frac{2}{t}} - 1}{1 + A e^{-\frac{2}{t}}}
where A is the constant of integration.
2.5.23
Problem:
Solve the differential equation:
y - t \frac{dy}{dt} = 6 - 3t^2 \frac{dy}{dt}
Solution:
Rearrange the equation:
Start by moving all terms involving \frac{dy}{dt} to one side:
y - t \frac{dy}{dt} = 6 - 3t^2 \frac{dy}{dt}
Rearranging the terms involving \frac{dy}{dt}:
y = 6 + \frac{dy}{dt} \left( 3t^2 - t \right)
Combine the results from both sides:
\ln|y - 6| = \ln|t| - \ln|1 - 3t| + C
Simplify the logarithms:
Simplify the right-hand side using properties of logarithms:
\ln|y - 6| = \ln\left|\frac{t}{1 - 3t}\right| + C
Exponentiate both sides to remove the logarithms:
|y - 6| = e^C \left|\frac{t}{1 - 3t}\right|
Let A = e^C, so:
y - 6 = A \frac{t}{1 - 3t}
Solve for y:
y = 6 + A \frac{t}{1 - 3t}
Final Answer:
The solution to the differential equation is:
y = 6 + A \frac{t}{1 - 3t}
where A is the constant of integration.
2.5.29
Problem:
Solve the initial value problem y' = 10y, with the initial condition y(0) = 3.
Solution:
We know from Problem 2.5.15 that the general solution to the differential equation y' = 10y is:
y = A e^{10t}
Use the initial condition y(0) = 3 to find A:
y(0) = A e^{10 \cdot 0} = A = 3
Therefore, A = 3.
The particular solution is:
y = 3 e^{10t}
Plotting the Direction Field and Solution:
Show code
import numpy as npimport matplotlib.pyplot as plt# Define the differential equation dy/dt = 10ydef dydt(t, y):return10* y# Generate grid for direction fieldt = np.linspace(0, 0.5, 20) y = np.linspace(-5, 30, 20) T, Y = np.meshgrid(t, y)# Calculate slope for direction fieldslope = dydt(T, Y)# Normalize the direction field arrowsnorm = np.sqrt(1+ slope**2)T_dir =1/ normY_dir = slope / norm# Plot the direction fieldplt.figure(figsize=(8, 6))plt.quiver(T, Y, T_dir, Y_dir, angles="xy", color="gray")# Plot the solution curve t_vals = np.linspace(0, 0.5, 200)y_vals =3* np.exp(10* t_vals)plt.plot(t_vals, y_vals, 'r', linewidth=2, label="Particular Solution")# Add labels, title, and limitsplt.title("Direction Field and Particular Solution", fontsize=14)plt.xlabel("t", fontsize=12)plt.ylabel("y", fontsize=12)plt.xlim([0, 0.5])plt.ylim([0, 30])# Add grid and legendplt.grid(True)plt.legend(loc="upper left")# Display the plotplt.show()
Final Answer:
The solution to the initial value problem is:
y = 3 e^{10t}
This solution represents exponential growth with an initial value of 3 at t = 0, as reflected in the direction field and solution curve plot.
2.5.37
Problem:
Solve the initial value problem:
y - t \frac{dy}{dt} = 6 - 3t^2 \frac{dy}{dt}, \quad y(1) = 5
Solution:
We know from Problem 2.5.23 that the general solution to this differential equation is:
y = 6 + A \frac{t}{1 - 3t}
Use the initial condition y(1) = 5 to find the value of A.
Substitute the initial condition t = 1 and y = 5 into the general solution:
5 = 6 + A \frac{1}{1 - 3 \cdot 1}
Simplify the equation:
5 = 6 + A \frac{1}{-2}
This simplifies to:
5 = 6 - \frac{A}{2}
Solve for A:
Subtract 6 from both sides:
-1 = -\frac{A}{2}
Multiply both sides by -2:
A = 2
Substitute A = 2 into the general solution:
y = 6 + 2 \frac{t}{1 - 3t}
The particular solution to the initial value problem is:
y = 6 + \frac{2t}{1 - 3t}
Plotting the Direction Field and Solution:
Show code
# Define the differential equation dy/dt = (y - 6) / (t - 3t^2)def dydt(t, y):return (y -6) / (t - (3* t**2))# Generate grid for direction fieldt = np.linspace(0.5, 5, 20) y = np.linspace(5, 6, 20) T, Y = np.meshgrid(t, y)# Calculate slope for direction fieldslope = dydt(T, Y)# Normalize the direction field arrowsnorm = np.sqrt(1+ slope**2)T_dir =1/ normY_dir = slope / norm# Plot the direction fieldplt.figure(figsize=(8, 6))plt.quiver(T, Y, T_dir, Y_dir, angles="xy", color="gray")# Plot the particular solution curve y = 6 + 2t / (1 - 3t)t_vals = np.linspace(0.5, 5, 200) y_vals =6+ ((2* t_vals) / (1-3* t_vals))plt.plot(t_vals, y_vals, 'r', linewidth=2, label="Particular Solution")# Add labels, title, and limitsplt.title("Direction Field and Particular Solution", fontsize=14)plt.xlabel("t", fontsize=12)plt.ylabel("y", fontsize=12)plt.xlim([0.5, 5]) plt.ylim([5, 6]) # Add grid and legendplt.grid(True)plt.legend(loc="upper left")# Display the plotplt.show()
Final Answer:
The solution to the initial value problem is:
y = 6 + \frac{2t}{1 - 3t}
2.5.41
Problem:
Solve the initial value problem:
(y + t) y' + y = t, \quad y(0) = 1
with the initial condition y(0) = 1, and then plot an appropriate direction field and sketch your solution.
Solution:
Rearrange the equation
The given equation is:
(y + t) \frac{dy}{dt} + y = t
Rewrite it as:
\frac{dy}{dt} = \frac{t - y}{y + t}
Cross multiply:
(t-y)dt = (y+t)dy
This can be rewritten as:
(y - t) dt + (y + t) dy = 0
Which is now in the standard form of an exact equation:
M(t, y) dt + N(t, y) dy = 0
Identify M(t, y) and N(t, y)
From the rewritten form of the equation:
(y - t) dt + (y + t) dy = 0
we can identify the functions M(t, y) and N(t, y) as:
M(t, y) = y - t \quad \text{and} \quad N(t, y) = y + t
Test for exactness
To check if the equation is exact, verify the condition for exactness:
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}
Compute the partial derivatives:
M(t, y) = y - t, so \frac{\partial M}{\partial y} = 1
N(t, y) = y + t, so \frac{\partial N}{\partial t} = 1
Since \frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}, the equation is exact.
Find the potential function
Since the equation is exact, there exists a potential function \Phi(t, y) such that:
\frac{\partial \Phi}{\partial t} = M(t, y) = y - t
\quad \text{and} \quad
\frac{\partial \Phi}{\partial y} = N(t, y) = y + t
Integrate M(t, y) with respect to t
Integrate M(t, y) = y - t with respect to t:
\Phi(t, y) = \int (y - t) \, dt = yt - \frac{t^2}{2} + h(y)
where h(y) is a function of y only.
Differentiate \Phi(t, y) with respect to y
Differentiate \Phi(t, y) with respect to y and set it equal to N(t, y) = y + t:
\frac{\partial \Phi}{\partial y} = t + h'(y) = y + t
Simplifying:
h'(y) = y
Integrate h'(y)
Integrating h'(y) = y with respect to y:
h(y) = \frac{y^2}{2} + C
The potential function is:
\Phi(t, y) = yt - \frac{t^2}{2} + \frac{y^2}{2} + C
General solution
The general solution of an exact differential equation is obtained by setting the potential function \Phi(t, y) equal to a constant:
yt - \frac{t^2}{2} + \frac{y^2}{2} = K
Apply the initial condition
We are given y(0) = 1. Substituting t = 0 and y = 1 into the equation:
1(0) - \frac{0^2}{2} + \frac{1^2}{2} = K
K = \frac{1}{2}
Final implicit solution
Substitute K = \frac{1}{2} back into the equation:
yt - \frac{t^2}{2} + \frac{y^2}{2} = \frac{1}{2}
Multiply through by 2 to simplify:
2yt - t^2 + y^2 = 1
This is the implicit solution of the differential equation:
2yt - t^2 + y^2 = 1
Solve for y using the quadratic formula
Solve this quadratic equation for y. Rearrange the equation:
y^2 + 2ty + (- t^2 - 1) = 0
This is a quadratic equation of the form ay^2 + by + c = 0, where:
a = 1
b = 2t
c = - t^2 - 1
Use the quadratic formula to solve for y:
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Substitute the values of a, b, and c:
y = \frac{-2t \pm \sqrt{(2t)^2 - 4(1)(- t^2 - 1)}}{2}
Simplifying:
y = \frac{-2t \pm \sqrt{4t^2 - 4(- t^2 - 1)}}{2}
y = \frac{-2t \pm \sqrt{4t^2 + 4 + 4t^2}}{2}
y = \frac{-2t \pm \sqrt{8t^2 + 4}}{2}
Factor out 4 from inside the square root:
y = \frac{-2t \pm \sqrt{4(2t^2 + 1)}}{2}
y = \frac{-2t \pm 2\sqrt{2t^2 + 1}}{2}
y = -t \pm \sqrt{2t^2 + 1}
Apply the initial condition
We are given y(0) = 1. Substituting t = 0 into the equation:
1 = -0 \pm \sqrt{2(0)^2 - 1}
1 = \pm \sqrt{1}
1 \ne -1
So our only solution is:
y = -t + \sqrt{2t^2 - 1}
Plotting the Direction Field and Solution:
Show code
# Define the differential equation dy/dt = (t - y) / (y + t)def dydt(t, y):return (t - y) / (y + t)# Generate grid for direction fieldt = np.linspace(0.1, 5, 20) y = np.linspace(0.1, 5, 20) T, Y = np.meshgrid(t, y)# Calculate slope for direction fieldslope = dydt(T, Y)# Normalize the direction field arrowsnorm = np.sqrt(1+ slope**2)T_dir =1/ normY_dir = slope / norm# Plot the direction fieldplt.figure(figsize=(8, 6))plt.quiver(T, Y, T_dir, Y_dir, angles="xy", color="gray")# Define the particular solution y = -t + sqrt(2t^2 + 1)def particular_solution(t):return-t + np.sqrt(2* t**2+1)# Plot the particular solution curvet_vals = np.linspace(0, 5, 200) y_vals = particular_solution(t_vals)plt.plot(t_vals, y_vals, 'r', linewidth=2, label="Particular Solution")# Add labels, title, and limitsplt.title("Direction Field and Particular Solution", fontsize=14)plt.xlabel("t", fontsize=12)plt.ylabel("y", fontsize=12)plt.xlim([0, 5])plt.ylim([0, 5])# Add grid and legendplt.grid(True)plt.legend(loc="upper left")# Display the plotplt.show()
Final Answer:
The solution to the initial value problem is:
y = -t + \sqrt{2t^2 - 1}
SECTION 2.6
Assigned: 5, 7, 11
2.6.5
Problem:
Solve the initial value problem:
y' + 2ty = 0, \quad y(0) = -2
using Euler’s method with h = 0.1 on the interval [0, 1]. Additionally, find the exact solution and compare the values and plots of the approximate and exact solutions.
Solution:
Exact solution
The given equation is:
y' + 2ty = 0
This is a first-order linear differential equation, and it can be solved by using an integrating factor. The general form of a linear differential equation is:
y' + p(t) y = 0
where p(t) = 2t. The integrating factor \mu(t) is:
\mu(t) = e^{\int p(t) \, dt} = e^{t^2}
Multiplying both sides of the equation by e^{t^2}:
e^{t^2} y' + 2t e^{t^2} y = 0
This simplifies to:
\frac{d}{dt} \left( e^{t^2} y \right) = 0
Integrating both sides:
e^{t^2} y = C
The general solution is:
y = C e^{-t^2}
Apply the initial condition y(0) = -2:
-2 = C e^0 \quad \Rightarrow \quad C = -2
The particular solution is:
y = -2 e^{-t^2}
Euler’s method
Euler’s method uses the formula:
y_{n+1} = y_n + h f(t_n, y_n)
where f(t, y) = -2ty from the original equation y' = -2ty.
We are given h = 0.1 and the initial condition y(0) = -2. We will approximate the solution for t \in [0, 1].
Step size: h = 0.1
Initial condition: y(0) = -2
Differential equation: y' = -2ty
Comparison of approximate and exact solutions
Table:
Show code
import pandas as pd# Given initial condition and step sizeh =0.1t_vals_euler = np.arange(0, 1.1, h)y_euler = np.zeros(len(t_vals_euler))y_euler[0] =-2# y(0) = -2# Define the function for the differential equation y' = -2tydef f(t, y):return-2* t * y# Euler's method iterationfor i inrange(1, len(t_vals_euler)): y_euler[i] = y_euler[i-1] + h * f(t_vals_euler[i-1], y_euler[i-1])# Exact solutiondef exact_solution(t):return-2* np.exp(-t**2)# Create a table for Euler's method and the exact solutiont_vals_exact_small = t_vals_euler # Use the same t-values for comparisony_exact_small = exact_solution(t_vals_exact_small)# Create a DataFrame to compare Euler's method with the exact solutiondata = {'t': t_vals_euler,'Euler Approximation': y_euler,'Exact Solution': y_exact_small}df_comparison = pd.DataFrame(data)table_markdown = df_comparison.to_markdown(index=False)# print(table_markdown)
t
Euler Approximation
Exact Solution
0
-2
-2
0.1
-2
-1.9801
0.2
-1.96
-1.92158
0.3
-1.8816
-1.82786
0.4
-1.7687
-1.70429
0.5
-1.62721
-1.5576
0.6
-1.46449
-1.39535
0.7
-1.28875
-1.22525
0.8
-1.10832
-1.05458
0.9
-0.930992
-0.889716
1
-0.763413
-0.735759
Graph:
Show code
t_vals_exact = np.linspace(0, 1, 200)y_exact = exact_solution(t_vals_exact)# Plot Euler's Method approximation and exact solutionplt.figure(figsize=(8, 6))plt.plot(t_vals_exact, y_exact, 'r', label="Exact solution", linewidth=2)plt.plot(t_vals_euler, y_euler, 'bo-', label="Euler's Method", markersize=5)# Add labels, title, and legendplt.title("Euler's Method vs Exact Solution", fontsize=14)plt.xlabel("t", fontsize=12)plt.ylabel("y", fontsize=12)plt.grid(True)plt.legend()# Display the plotplt.show()
Final Answer:
The exact solution to the differential equation is:
y = -2 e^{-t^2}
The graph compares Euler’s method with the exact solution. Euler’s method provides a reasonable approximation but slightly underestimates the true values as t increases.
2.6.7
Problem:
Solve the initial value problem using Euler’s method with h = 0.1 on the interval [0, 1]:
y' - y = 0, \quad y(0) = 2
In addition, find the exact solution and compare the values and plots of the approximate and exact solutions.
Solution:
Exact Solution
The given equation is:
y' - y = 0
This is a separable differential equation. Rearrange it:
\frac{dy}{dt} = y
Separate the variables:
\frac{1}{y} \, dy = dt
Integrate both sides:
\int \frac{1}{y} \, dy = \int dt
\ln|y| = t + C
Exponentiate both sides to solve for y:
|y| = e^{t+C} = e^C \cdot e^t
Let A = e^C, so:
y = A e^t
Use the initial condition y(0) = 2 to find A:
2 = A e^0 \quad \Rightarrow \quad A = 2
The particular solution is:
y = 2 e^t
Euler’s Method
Euler’s method uses the formula:
y_{n+1} = y_n + h f(t_n, y_n)
where f(t, y) = y (from the original equation y' = y).
# Plot Euler's Method approximation and exact solutionplt.figure(figsize=(8, 6))plt.plot(t_vals_exact, y_exact, 'r', label="Exact Solution", linewidth=2)plt.plot(t_vals_euler, y_euler, 'bo-', label="Euler's Method", markersize=5)# Add labels, title, and legendplt.title("Euler's Method vs Exact Solution", fontsize=14)plt.xlabel("t", fontsize=12)plt.ylabel("y", fontsize=12)plt.grid(True)plt.legend()# Display the plotplt.show()
Final Answer:
The exact solution to the differential equation is:
y = 2 e^t
The graph compares Euler’s method with the exact solution. Euler’s method provides a reasonable approximation but slightly underestimates the true values as t increases.
2.6.11
Problem:
Solve the initial value problem using Euler’s method with h = 0.1 on the interval [0, 1]:
(y')^2 - 2y^2 = t, \quad y(0) = 2
In addition, explain why it is not possible to solve the IVP exactly by established methods.
Solution:
Rearrange the Differential Equation
We are given:
(y')^2 - 2y^2 = t
First, solve for y':
(y')^2 = t + 2y^2
Taking the square root of both sides (assuming the positive root):
y' = \sqrt{t + 2y^2}
Euler’s Method
Euler’s method uses the formula:
y_{n+1} = y_n + h f(t_n, y_n)
where f(t, y) = \sqrt{t + 2y^2}, and the initial condition is y(0) = 2.
Given:
Step size h = 0.1
Initial condition y(0) = 2
Differential equation: y' = \sqrt{t + 2y^2}
Calculate the approximate solution using Euler’s method.
Table
Show code
h =0.1t_vals_euler = np.arange(0, 1.1, h)y_euler = np.zeros(len(t_vals_euler))y_euler[0] =2# Initial condition y(0) = 2# Define the differential equation y' = sqrt(t + 2y^2)def f(t, y):return np.sqrt(t +2* y**2)# Euler's method iterationfor i inrange(1, len(t_vals_euler)): y_euler[i] = y_euler[i-1] + h * f(t_vals_euler[i-1], y_euler[i-1])# Create a DataFrame to display the Euler method approximationdf_euler_approximation = pd.DataFrame({'t': t_vals_euler,'Euler Approximation': y_euler})table_markdown = df_euler_approximation.to_markdown(index=False)# print(table_markdown)
t
Euler Approximation
0
2
0.1
2.28284
0.2
2.60723
0.3
2.97865
0.4
3.40344
0.5
3.8889
0.6
4.4434
0.7
5.07655
0.8
5.79934
0.9
6.62435
1
7.56597
Graph
Show code
plt.figure(figsize=(8, 6))plt.plot(t_vals_euler, y_euler, 'bo-', label="Euler's Method", markersize=5)# Add labels, title, and legendplt.title("Euler's Method Approximation", fontsize=14)plt.xlabel("t", fontsize=12)plt.ylabel("y", fontsize=12)plt.grid(True)plt.legend()# Display the plotplt.show()
Final Answer:
The equation involves (y')^2, making it non-linear and not separable. As a result, standard methods like separation of variables or integrating factors cannot be applied, so numerical methods such as Euler’s method are needed to approximate the solution.
SECTION 2.7
Assigned: 1, 3, 7
2.7.1
Problem:
For a population P(t) that exhibits logistic growth according to the general model
\frac{dP}{dt} = kP \left(1 - \frac{P}{A}\right), \quad P(0) = P_0
where k and A are constants, solve the following:
Determine the values of P (in terms of A and k) for which P is an increasing function.
Sketch by hand the direction field for the differential equation, clearly indicating the role of the constant A in your sketch.
Determine the value(s) of P (in terms of A and k) for which P is increasing most rapidly, and justify your answer.
Solve the initial-value problem explicitly for P to show that
P(t) = \frac{A}{1 + Me^{-kt}}
and determine M in terms of A and P_0.
Solution:
Part (a)
To determine when P(t) is an increasing function, analyze the sign of \frac{dP}{dt}:
\frac{dP}{dt} = kP \left(1 - \frac{P}{A}\right)
Since k is a positive constant, \frac{dP}{dt} will be positive when P \left(1 - \frac{P}{A}\right) > 0.
P > 0 for population values (negative population is impossible).
The expression 1 - \frac{P}{A} > 0 implies P < A.
So, \frac{dP}{dt} > 0 when 0 < P < A. Therefore, P(t) is increasing for 0 < P < A.
Part (b)
Show Code
import numpy as npimport matplotlib.pyplot as plt# Constantsk =0.75# Growth rate constantA =100# Carrying capacitydef dP_dt(P):return k * P * (1- P / A)P_values = np.linspace(0, 120, 20) t_values = np.linspace(0, 10, 20) P_grid, t_grid = np.meshgrid(P_values, t_values)dP = dP_dt(P_grid)norm = np.sqrt(1+ dP**2)dt =1/ normdP = dP / normplt.figure(figsize=(10, 6))plt.quiver(t_grid, P_grid, dt, dP, angles='xy', scale=20, color="blue")plt.axhline(A, color="red", linestyle="-", linewidth=9, label="Carrying Capacity A")plt.xticks([])plt.yticks([])plt.xlabel("Time (t)")plt.ylabel("Population (P)")plt.title("Direction Field for Logistic Growth Equation")plt.legend()plt.grid(True)plt.show()
Part (c)
To find the value of P at which P is increasing most rapidly take the derivative of \frac{dP}{dt} with respect to P and set it to zero:
Differentiate with respect to P:
\frac{d}{dP}\left(kP \left(1 - \frac{P}{A}\right)\right) = k \left(1 - \frac{2P}{A}\right)
Set the derivative to zero to find the critical point:
1 - \frac{2P}{A} = 0
Solving for P:
P = \frac{A}{2}
P = \frac{A}{2} is the population level at which P(t) is increasing most rapidly.
Got it! If we let M = e^{-C} instead of \frac{1}{M}, then the equation becomes simpler. I’ll go through the steps again with this adjustment.
Solution (with adjusted Step 4)
Part (d)
\frac{dP}{dt} = kP \left(1 - \frac{P}{A}\right)
Separate variables:
\frac{1}{P \left(1 - \frac{P}{A}\right)} \, dP = k \, dt
Rewrite the left side using partial fractions:
\frac{1}{P \left(1 - \frac{P}{A}\right)} = \frac{1}{P} + \frac{1}{A - P}
\left( \frac{1}{P} + \frac{1}{A - P} \right) \, dP = k \, dt
Integrate both sides:
\int \left( \frac{1}{P} + \frac{1}{A - P} \right) \, dP = \int k \, dt
Solve the integrals and multiply by -1:
\ln |A - P| - \ln |P| = -kt - C
Combine the logarithms:
\ln \left| \frac{A - P}{P} \right| = -kt - C
Exponentiate both sides to solve for P:
\frac{A - P}{P} = e^{-kt - C} = M e^{-kt} \quad \text{(where $M = e^{-C}$)}
Solve for P:
A - P = P \cdot M e^{-kt}
Move terms involving P to one side:
P + P \cdot M e^{-kt} = A
Factor out P:
P (1 + M e^{-kt}) = A
Solve for P:
P = \frac{A}{1 + M e^{-kt}}
Apply the initial condition P(0) = P_0 to determine M:
P_0 = \frac{A}{1 + M}
Solving for M:
M = \frac{A - P_0}{P_0}
The explicit solution is:
P(t) = \frac{A}{1 + \left(\frac{A - P_0}{P_0}\right)e^{-kt}}
where A is the carrying capacity, k is the growth rate constant, and P_0 is the initial population.
Final Answer:
(a)P(t) is an increasing function for 0 < P < A.
(b) The direction field shows arrows pointing upwards for 0 < P < A, horizontal at P = A (the carrying capacity), and downwards for P > A.
(c) The value of P at which P is increasing most rapidly is P = \frac{A}{2}.
(d) The explicit solution to the logistic growth equation is:
P(t) = \frac{A}{1 + \left(\frac{A - P_0}{P_0}\right)e^{-kt}}
where A is the carrying capacity, k is the growth rate constant, and P_0 is the initial population.
2.7.3
Problem:
Consider the differential equation
\frac{dP}{dt} = -0.02 P^2 + 0.08 P
What are the equilibrium solutions to this equation?
Determine whether each equilibrium solution is stable or unstable.
At what value of P is the function growing most rapidly?
Under the initial condition P(0) = 0.25, determine the time at which P(t) = 3.
Solution:
Part (a)
To find the equilibrium solutions, set \frac{dP}{dt} = 0:
-0.02 P^2 + 0.08 P = 0
Factor out P:
P \left(-0.02 P + 0.08\right) = 0
Set each factor to zero:
P = 0
-0.02 P + 0.08 = 0
Solve for P in the second equation:
-0.02 P = -0.08
P = 4
The equilibrium solutions are P = 0 and P = 4.
Part (b)
To determine the stability of each equilibrium solution, examine the sign of \frac{dP}{dt} around P = 0 and P = 4.
For P = 0:
When P is slightly positive (just above 0), \frac{dP}{dt} = -0.02 P^2 + 0.08 P is positive, which means P increases.
Therefore, P = 0 is unstable.
For P = 4:
When P is slightly less than 4, \frac{dP}{dt} = -0.02 P^2 + 0.08 P is positive, which means P increases toward 4.
When P is slightly more than 4, \frac{dP}{dt} is negative, which means P decreases toward 4.
Therefore, P = 4 is stable.
Part (c)
Differentiate \frac{dP}{dt} with respect to P:
\frac{d}{dP} \left(-0.02 P^2 + 0.08 P\right) = -0.04 P + 0.08
Set the derivative to zero to find the critical point:
-0.04 P + 0.08 = 0
P = 2
P(t) is growing most rapidly when P = 2.
Part (d)
To solve for the time t when P(t) = 3 with the initial condition P(0) = 0.25, we start by solving the differential equation explicitly.
Rewrite the differential equation:
\frac{dP}{dt} = -0.02 P^2 + 0.08 P
Factor the right side:
\frac{dP}{dt} = 0.02 P (4 - P)
Use partial fraction decomposition on the left side:
\frac{1}{P (4 - P)} = \frac{1}{4} \left(\frac{1}{P} + \frac{1}{4 - P}\right)
So the equation becomes:
\frac{1}{4} \left(\frac{1}{P} + \frac{1}{4 - P}\right) \, dP = 0.02 \, dt
Integrate both sides:
\int \frac{1}{4} \left(\frac{1}{P} + \frac{1}{4 - P}\right) \, dP = \int 0.02 \, dt
\frac{1}{4} \left(\ln |P| - \ln |4 - P|\right) = 0.02 t + C
Combine the logarithms:
\ln \left| \frac{P}{4 - P} \right| = 0.08 t + B
where B = 4C is a constant.
Exponentiate both sides to solve for P:
\frac{P}{4 - P} = M e^{0.08 t}
where M = e^{B}.
Apply the initial condition P(0) = 0.25 to find M:
\frac{0.25}{4 - 0.25} = M
M = \frac{0.25}{3.75} = \frac{1}{15}
Solve for t when P(t) = 3:
\frac{3}{4 - 3} = \frac{1}{15} e^{0.08 t}
3 = \frac{1}{15} e^{0.08 t}
e^{0.08 t} = 45
0.08 t = \ln 45
t = \frac{\ln 45}{0.08}
Final Answer:
(a) The equilibrium solutions are P = 0 and P = 4.
(b) The equilibrium P = 0 is unstable, and P = 4 is stable.
(c) The function P(t) is growing most rapidly when P = 2.
(d) The time t at which P(t) = 3 with the initial condition P(0) = 0.25 is:
t = \frac{\ln 45}{0.08}
2.7.7
Problem:
A cylindrical tank of height 4 \, \text{m} and radius 1 \, \text{m} is full of water. A small hole of diameter 1 \, \text{cm} is opened in the bottom of the tank. Use Torricelli’s law to determine how long it will take for all the water to drain from the tank.
Solution:
Torricelli’s Law:
\frac{dV}{dt} = -a v = -a \sqrt{2gh}
Where:
V is the volume of water in the tank,
a is the cross-SECTIONal area of the hole,
v = \sqrt{2gh} is the velocity of the water leaving the hole,
g is the acceleration due to gravity (9.8 \, \text{m/s}^2),
h is the height of the water above the hole.
Define Areas
Let:
The cross-SECTIONal area of the tank be A_t = \pi \times (1)^2 = \pi \, \text{m}^2,
The area of the hole be a = \pi \times \left(\frac{0.01}{2}\right)^2 = 5 \times 10^{-5} \pi \, \text{m}^2.
Relate \frac{dV}{dt} and \frac{dh}{dt}
The volume V of water in the tank is related to the height h by V = A_t h.
\frac{dV}{dt} = A_t \frac{dh}{dt}
Substitute this into our initial equation:
A_t \frac{dh}{dt} = -a \sqrt{2gh}
Integrate both sides with respect to h and t:
\int \frac{1}{\sqrt{h}} \, dh = -2.5 \times 10^{-5} \sqrt{2g} \int dt
2 \sqrt{h} = -2.5 \times 10^{-5} \sqrt{2g} \, t + C
Apply Initial Conditions
Initially, h(0) = 4 meters. Substitute h = 4 and t = 0 to solve for C:
2 \sqrt{4} = C
C = 4
So the equation becomes:
2 \sqrt{h} = -2.5 \times 10^{-5} \sqrt{2g} \, t + 4
Solve for t when h = 0
When the tank is empty, h = 0. Substitute h = 0 into the equation:
2 \sqrt{0} = -2.5 \times 10^{-5} \sqrt{2g} \, t + 4
0 = -2.5 \times 10^{-5} \sqrt{2g} \, t + 4
2.5 \times 10^{-5} \sqrt{2g} \, t = 4
t = \frac{4}{2.5 \times 10^{-5} \sqrt{2g}}
Substitute g = 9.8 \, \text{m/s}^2 and Calculate t
t = \frac{4}{2.5 \times 10^{-5} \sqrt{2 \times 9.8}}
